RE: 0^0 = ?

[Randy Gober]

Thanks for the explanation John.

As a small fyi, L'Hopitals really doesn't apply. 
Since we need lim x->0, to apply L'Hopitals, you need both the numerator and
the denominator to be continuous on an interval [a,b] about zero and
differentiable on an interval (a,b) about zero. 

 ln(x) and 1/x are neither.


Thanks,

--Randy


-----Original Message-----
From: John W. Eaton [mailto:address@hidden 
Sent: Thursday, November 13, 2003 10:51 AM
To: address@hidden
Cc: address@hidden; 'Cong'
Subject: RE: 0^0 = ?


On 13-Nov-2003, address@hidden <address@hidden> wrote:

| I know that by L'Htpital's Rule you should get:
| 
| ln(y)=x*ln(x) = ln(x)/(1/x)  so
| 
| Lim x->0+  ln(x)/(1/x) = ( 1/x )/( -1/( x^2)) =
| 
| Lim x->0+ (-x) = 0
| so  ln(y) = 0 and then y=1.
| 
| Maybe this is the reason for the behavior?

The 0^0 == 1 behavior is part of the IEEE 754 standard for floating point
arithmetic.  The paper "What every computer scientist should know about
floating point arithmetic" by David Goldberg provides a rationale for the
behavior that is a bit different than above (it's at the end of a section
titled "ambiguity").  To start with, I think you need to look at this as
y^x, not x^x.  A quick google search should turn up a copy of the paper if
you want the details.

jwe



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